bzoj2337: [HNOI2011]XOR和路径

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2337

因为是xor，我们按位处理，可以得到如果权值(w&(1<<i))，那么ex[u]+=1/d[u]*(1-f[v])，否则ex[u]+=1/d[u]*f[v]，然后高斯消元。

注意几个点，终点连出去的边全部不用考虑，不用去判断权值w>0的情况，令a[n][n]=1，有重边就不要作死开邻接矩阵。

 
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 210
#define ll long long
#define eps 1e-6
using namespace std;
struct data{int obj,pre,c;
}e[200500];
int d[maxn],bin[maxn],head[maxn];
double a[maxn][maxn],ans;
int n,x,y,z,m,tot;
void insert(int x,int y,int z){
    d[x]++; e[++tot].obj=y; e[tot].pre=head[x]; e[tot].c=z; head[x]=tot;
}
ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
void solve(int n){
    int now; double t;
    rep(i,1,n){
        now=i;
        while (fabs(a[now][i])<=eps) now++;
        if (now>n) continue;
        if (now!=i) rep(j,1,n+1) swap(a[i][j],a[now][j]);
        t=a[i][i];
        rep(j,1,n+1) a[i][j]/=t;
        rep(j,1,n) if (j!=i){
            t=a[j][i];
            rep(k,1,n+1) a[j][k]-=t*a[i][k];
        } 
    }
}
int main(){
    bin[0]=1;
    rep(i,1,30) bin[i]=bin[i-1]*2;
    n=read(); m=read();
    rep(i,1,m){
        int x,y,z;
        x=read(); y=read(); z=read();
        insert(x,y,z);
        if (x!=y) insert(y,x,z);
    }
    ans=0;
    rep(o,0,30){
        clr(a,0);
        rep(u,1,n-1){
            a[u][u]=1.0;
            for (int j=head[u];j;j=e[j].pre){
                int v=e[j].obj;
                    if (e[j].c&bin[o]) a[u][n+1]+=1.0/d[u],a[u][v]+=1.0/d[u];
                    else a[u][v]-=1.0/d[u];
            }
        }   
        a[n][n]=1.0;
        solve(n);
        ans+=a[1][n+1]*bin[o];
    }   
    printf("%.3lf
",ans);
      
    return 0;
}